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I learned electrodynamics. According to the vector potential determination, $$ \mathbf B = [\nabla \times \mathbf A ], $$ Coulomb gauge, $$ \nabla \mathbf A = 0, $$ and one of Maxwell's equations, $$ [\nabla \times \mathbf B ] = \frac{1}{c}4\pi \mathbf j, $$

I can assume, that

$$ [\nabla \times \mathbf B ] = \nabla (\nabla \mathbf A) - \Delta \mathbf A = -\Delta \mathbf A = \frac{1}{c}4 \pi \mathbf j. $$ How to prove that the one of the solutions of this equation is solution like newtonian potential, $$ \mathbf A = \frac{1}{c}\int \limits_{V} \frac{\mathbf j (r) d^{3}\mathbf r}{|\mathbf r - \mathbf r_{0}|}? $$

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It's exactly the same proof as for scalar solution $\phi = \int\rho/r$. You just rename $\phi$ as $A_x$ or $A_y$ or $A_z$ and you rename $\rho$ as $j_x,j_y,j_z$ and the solution of the same Laplace equation is obviously the same, too. What's the problem here? It's the same Laplace equation for a triplet of quantities - three equal Laplace equations. If you can't derive the Coulomb's law from the Laplace equation, please post it as a new question. –  Luboš Motl Jun 14 '12 at 11:08
    
Well, noticing that every component of the equation $\Delta \mathbf{A} = -\frac{4 \pi}{c} \mathbf{j}$ satisfies Poisson-type equation, we immediately write down the solution. UPDATE: Luboš wrote the same thing above =) –  Physicsworks Jun 14 '12 at 11:14

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